Current Location  >  Physics Formulas > Electromagnetism > Electric Field Due to a System of Discrete Charges

Electric Field Due to a System of Discrete Charges

The electric field or electric field strength is the electrostatic force acting on a small positive test charge placed at that point. If is the electrostatic force experienced by a test charge q at a point, then the electric field intensity at that point is given by

S.I unit of electric field intensity is Newton/coulomb (NC-1).

If the test charge is not small, then the electric field may be affected by the test charge and hence we modify the above equation as follows:

Consider a system of charges q1, q2, ………..qn placed at distances r1, r2….rn with respect to some origin. Then the electric field intensity due to all these charges at a point is found out using the Principle of superposition. Let  intensity due to the number of charges q1, q2, ………..qn. Then the resultant electric intensity  at that point due to these charges is given by the superposition theorem.

Electric field intensity due to the nth charge is

Magnitude of the electric field intensity is given by the equation:

Example1: Two point charges of 1μC and -1 μC are separated by a distance of 100 Å. A point P is at a distance of 10 cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at P will be

(a)    9 N/C

(b)   0.9 V/m

(c)    90 V/m

(d)   0.09 N/C

Solution: The point lies on equatorial line of a short dipole. So, 

Example2: Three charges 2q,-q and –q are located at the vertices of an equilateral triangle .At the center of the triangle

(a)    the field is zero but potential is non-zero.

(b)   the field is non-zero ,but potential is zero.

(c)    both field and potential are zero.

(d)   both field and potential are non zero.

Solution: (b)

Obviously, E ≠ 0.Hence the field is non-zero but potential is zero.

Example3: ABC is an equilateral triangle. Charges + q are placed at each corner. As shown in figure. The electric intensity at centre O will be

Solution: Unit Positive charge at O will be repelled equally by three charges at the three corners of triangle

By symmetry, resultant at O would be zero. © 2017 | Contact us | Terms of Use | Privacy Policy | Yellow Sparks Network
Web Formulas